Evaluation of Definite Integrals by Substitution - Practice Questions & MCQ

We have already learned to find Indefinite Integration by using the substitution method. But in the case of definite integration, we also need to change the limits of integration 'a' and 'b'. If we substitute x = g(t), then g(t)  must be continuous in the interval [a, b].

Let's look at some examples of how such questions are solved.

Example 1

Compute the integral $\int_0^{\pi / 2} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}$
Let $\quad I=\int_{x=0}^{x=\pi / 2} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}$
Divide numerator and denominator by $\cos ^2 x$
$
=\int_{x=0}^{x=\pi / 2} \frac{\sec ^2 x d x}{a^2+b^2 \tan ^2 x}
$

Put $\quad \tan x=t \Rightarrow \sec ^2 x d x=d t$
$
\therefore \quad I=\int_{t=0}^{t=\infty} \frac{d t}{a^2+b^2 t^2}
$

We find the new limits of integration $t=\tan x \Rightarrow t=0$ when $x=0$ and $t=\infty$ when $x=\pi / 2$
$
\begin{aligned}
\Rightarrow \quad I & =\frac{1}{b^2} \int_0^{\infty} \frac{d t}{\left(\frac{a}{b}\right)^2+t^2}=\frac{1}{b^2} \cdot \frac{1}{a / b}\left[\tan ^{-1} \frac{b t}{a}\right]_0^{\infty} \\
& =\frac{1}{a b}\left[\frac{\pi}{2}-0\right]=\frac{\pi}{2 a b}
\end{aligned}
$